Concept-wise Practice

reciprocal-sum MCQ Questions for Class 10

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Practice Questions

13 questions tagged with reciprocal-sum.

समीकरण \(3x^2-10x+3=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(3x^2-10x+3=0\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{10}{3}\)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{\frac{10}{3}}{1}=\frac{10}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{10}{3}\). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{\frac{10}{3}}{1}=\frac{10}{3}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहां \(\frac{\frac{10}{3}}{1}=\frac{10}{3}\) है।

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समीकरण \(2x^2-7x+3=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(2x^2-7x+3=0\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{7}{3}\)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{7}{3}\). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{3}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहां \(\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{3}\) है।

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यदि द्विघात समीकरण के मूल (4) और (-9) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?

If the roots of a quadratic equation are (4) and (-9), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{5}{36}\)

Step 1

Concept

Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{5}{36}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).

Step 3

Exam Tip

यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-36\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\) है।

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यदि द्विघात समीकरण के मूल (2) और (-7) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?

If the roots of a quadratic equation are (2) and (-7), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{5}{14}\)

Step 1

Concept

Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{5}{14}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).

Step 3

Exam Tip

यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\) है।

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यदि \(\alpha\) और \(\beta\) समीकरण \(3x^2-4x-4=0\) के मूल हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या है?

If \(\alpha\) and \(\beta\) are roots of \(3x^2-4x-4=0\), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. (-1)

Step 1

Concept

Here \(\alpha+\beta=\frac{4}{3}\) and \(\alpha\beta=-\frac{4}{3}\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-1\).

Step 2

Why this answer is correct

The correct answer is A. (-1). Here \(\alpha+\beta=\frac{4}{3}\) and \(\alpha\beta=-\frac{4}{3}\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-1\).

Step 3

Exam Tip

यहां \(\alpha+\beta=\frac{4}{3}\) और \(\alpha\beta=-\frac{4}{3}\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-1\) होगा।

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यदि द्विघात समीकरण के मूल (3) और (-4) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?

If the roots of a quadratic equation are (3) and (-4), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. -\(\frac{1}{12}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).

Step 2

Why this answer is correct

The correct answer is A. -\(\frac{1}{12}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{-1}{-12}\) नहीं बल्कि \(-\frac{1}{12}\) क्योंकि योग (-1) और गुणनफल (-12) है।

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समीकरण \(6x^2-19x+10=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(6x^2-19x+10=0\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{19}{10} \)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{19}{6}}{\frac{10}{6}}=\frac{19}{10}\).

Step 2

Why this answer is correct

The correct answer is A. \( \frac{19}{10} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{19}{6}}{\frac{10}{6}}=\frac{19}{10}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{19}{6}}{\frac{10}{6}}=\frac{19}{10}\) है।

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समीकरण \(5x^2-17x+6=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(5x^2-17x+6=0\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{17}{6} \)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{17}{5}}{\frac{6}{5}}=\frac{17}{6}\).

Step 2

Why this answer is correct

The correct answer is A. \( \frac{17}{6} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{17}{5}}{\frac{6}{5}}=\frac{17}{6}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{17}{5}}{\frac{6}{5}}=\frac{17}{6}\) है।

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समीकरण \(4x^2-13x+9=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(4x^2-13x+9=0\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{13}{9} \)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}\).

Step 2

Why this answer is correct

The correct answer is A. \( \frac{13}{9} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{13}{4}}{\frac{9}{4}}=\frac{13}{9}\) है।

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समीकरण \(3x^2-10x+7=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(3x^2-10x+7=0\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{10}{7} \)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\).

Step 2

Why this answer is correct

The correct answer is A. \( \frac{10}{7} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}\) है।

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समीकरण \(2x^2-5x+3=0\) के मूलों के व्युत्क्रमों का योग क्या है?

What is the sum of reciprocals of the roots of \(2x^2-5x+3=0\)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{5}{3} \)

Step 1

Concept

The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\).

Step 2

Why this answer is correct

The correct answer is A. \( \frac{5}{3} \). The sum of reciprocals is \(\frac{\alpha+\beta}{\alpha\beta}\). Here it is \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\).

Step 3

Exam Tip

व्युत्क्रमों का योग \(\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ \(\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{3}\) है।

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यदि \(\alpha=2+\sqrt{5}\) और \(\beta=2-\sqrt{5}\), तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या है?

If \(\alpha=2+\sqrt{5}\) and \(\beta=2-\sqrt{5}\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. (-4)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\). Finding sum and product first is easier.

Step 2

Why this answer is correct

The correct answer is A. (-4). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\). Finding sum and product first is easier.

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4\)। पहले योग और गुणनफल निकालना आसान रहता है।

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यदि \(\alpha=5+\sqrt{6}\) और \(\beta=5-\sqrt{6}\), तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) क्या है?

If \(\alpha=5+\sqrt{6}\) and \(\beta=5-\sqrt{6}\), what is \(\frac{1}{\alpha}+\frac{1}{\beta}\)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{10}{19}\)

Step 1

Concept

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (10) and product is (25-6=19), so the answer is \(\frac{10}{19}\).

Step 2

Why this answer is correct

The correct answer is A. \(\frac{10}{19}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (10) and product is (25-6=19), so the answer is \(\frac{10}{19}\).

Step 3

Exam Tip

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) होता है। यहाँ योग (10) और गुणनफल (25-6=19), इसलिए उत्तर \(\frac{10}{19}\) है।

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